Effects of capacitors on potential difference

AI Thread Summary
The discussion centers on comparing the potential differences across two parallel-plate capacitors with different plate separations and charge quantities. Capacitor 1 has a plate separation twice that of capacitor 2 and holds six times the charge. The initial calculation suggested a potential difference ratio of 3, but this was marked incorrect in MasteringPhysics. The confusion arose from misunderstanding the relationship between capacitance, charge, and potential difference, particularly how capacitance is inversely proportional to plate separation. Ultimately, the participants recognized the need to carefully apply the equations governing capacitance and potential difference to arrive at the correct conclusion.
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Homework Statement


Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.

How do the potential differences across each of the two capacitors compare to each other? (V1/V2 = ?)

Homework Equations



I tried two different approaches:

V(A->B) = -W(A->B)/qtest = -(integral of E*dl from A to B)

V = q/C
C = εA/d

The Attempt at a Solution



q1 = 6q2
r1 = 2q2

V1/V2 = (kq1/r1^2)*r1 / (kq2/r2^2)*r^2 = (6q2/2r2)/(q2/r2) = 3

This is not correct.

C1/C2 = d1/d2 = 2

V = q/C = 6/2 = 3

I must be misunderstanding potential difference but I'm not sure where I'm going wrong.
 
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? Why do you think 3 is wrong ?
 
Cutter Ketch said:
? Why do you think 3 is wrong ?

Is it not wrong? When I tried the answer on MasteringPhysics it was marked as incorrect.
 
Well it's pretty straight forward. V is proportional to q and inversely proportional to d, and that is all that changed. Hard to get a different answer.
 
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$
 
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Oh, I suck! I didn't even look at the equations. It just seemed so obvious and natural that the potential difference MUST get bigger as the plates get closer. You know, it seemed so obvious that I'm having trouble getting my head back on straight.
 
Sorry about that :smile:
 
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Wow, I can't believe I messed that up. Thank you!
 
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